控制檯中的測試

Question

諮詢?- go(c, g).返回false，但對於?- go(a, d).表示正確。我實際上並不瞭解它，因爲我已經添加了適當的規則，並且對於其中的大多數規則而言都是有效的。

下面是我使用的語句：

door(a, b). 
door(b, c). 
door(c, d). 
door(b, e). 
door(e, f). 
door(e, g). 

go(FromRoom, ToRoom):- 
    door(FromRoom,ToRoom). 

go(FromRoom, ToRoom):- 
    door(ToRoom, FromRoom). 

go(FromRoom, ToRoom) :- 
    door(FromRoom, NextRoom), 
    go(NextRoom, ToRoom), !. 

go(FromRoom, ToRoom):- 
    door(ToRoom,NextRoom), 
    go(NextRoom, FromRoom), !. 

Answer 1

您需要蓄能器添加到您的謂詞，以避免卡在INF循環。

door(a, b). 
door(b, c). 
door(c, d). 
door(b, e). 
door(e, f). 
door(e, g). 

go(From, To, T) :- 
    T=[From|T1], 
    go(From, To, [To], T1). 
go(From, To,T, T) :- 
    door(From, To),!. 
go(From,To,T,T) :- 
    door(To,From),!. 
go(From, To, Acc,T) :- 
    door(X, Do),!, 
    \+ member(X, Acc), 
    go(From, X, [X|Acc],T). 
go(From,To,Acc,T) :- 
    door(X,Z),!, 
    \+member(Z,Acc), 
    go(X,To,[X|Acc],T). 

Answer 2

你第四條規則應該閱讀

go(FromRoom, ToRoom):- 
door(ToRoom,NextRoom), 
go(FromRoom,NextRoom),!. 

，因爲你已經在呼喚門/ 2交換搜索順序，但是這會導致循環，作爲正確的@whd說。這裏一個版本的 '回報' 的路徑

go(Room, Room, Path, Path). 

go(FromRoom, ToRoom, SoFar, Path) :- 
    (door(FromRoom, NextRoom) ; door(NextRoom, FromRoom)), 
    \+ member(NextRoom, SoFar), 
    go(NextRoom, ToRoom, [NextRoom|SoFar], Path). 

測試：

?- go(c,g,[],P). 
P = [g, e, b, c, d] ; 
P = [g, e, b] ; 
false. 

由於Prolog的搜索順序，這是需要一個旅遊'在d之前回去。很容易避免在頂層調用SoFar參數...